Question

ASAP! What is the y intercept of the line perpendicular to the line y = 4/3x + 1 that includes the point (4, 1)?

Answer 1

Answer: $x=\frac{16}{3}$

Step-by-step explanation:

The equation of the line in Slope-Intercept form is:

$y=mx+b$

Where "m" is the slope of the line and "b" is the y-intercept.

Knowing that the equation of the given line in Slope-Intercept form is:

$y = \frac{4}{3} x + 1$

You can identify that the slope of this line is:

$m=\frac{4}{3}$

Since the slope of perpendicular lines are negative recirprocals, we know that the slope of the other line perpendical to the line  $y = \frac{4}{3} x + 1$ is:

$m==-\frac{3}{4}[$

Now we need to substitute the slope and the  coordinates of the point (4,1) into $y=mx+b$ and solve for "b":

$1=-\frac{3}{4}(4)+b$

$1+3=b$

$b=4$

Then the equation of this line is:

 $y=-\frac{3}{4}x+4$

Finally, we must substitute $y=0$ into this equation and solve for "x":

$0=-\frac{3}{4}x+4\\\\-4=-\frac{3}{4}x\\\\(-4)(-4)=3x\\\\x=\frac{16}{3}$

Answer 2

Answer:

y-intercept is 4.

Step-by-step explanation:

Given,

Equation of the line,

$y=\frac{4}{3}x+1$

∵ Equation of a line is y = mx + c, where m is the slope of the line,

By comparing,

The slope of the above line is $\frac{4}{3}$,

Let m' is the slope of the line perpendicular to above line,

$\implies m'\times \frac{4}{3}=-1$

$\implies m' = -\frac{3}{4}$

Now, the equation of a line passes through a point $(x_1,y_1)$ is,

$y-y_1=m(x-x_1)$

So, the equation of the perpendicular line having slope $-\frac{3}{4}$ and passes through ( 4,1) is,

$y-1=-\frac{3}{4}(x-4)\implies y=-\frac{3}{4}x+3+1\implies y=-\frac{3}{4}x+4$

For y-intercept,

x = 0,

⇒ y = 4

Hence, the y-intercept of the perpendicular line is 4.